Question

If air core is replaced by an iron core in an inductor, its self-inductance is increased from $0.02mH$ to $40mH$. The relative permeability of iron is

• A. 5000
• B. 2000
• C. 200
• D. 500
• E. 400

Answer 1

Answer: (B)

Inductance of a solenoid coil of length $L$, total number of turns $N$ and cross-sectional area $A$, is given as
$L=\frac{{\mu }_0{N}^{2}A}{l}\dots \text{ (i) }$
where, ${\mu }_0$ is the permeability in vacuum (or air).
If air core is replaced by iron core, inductance will be given as
$L^{\prime }=\frac{\mu N^{2}A}{l}\dots \text{.(ii) }$
On dividing Eq. (ii) by Eq (i). we get
$\frac{L^{\prime }}{L}=\frac{\mu N^{2}A}{l}/\frac{{\mu }_0{N}^{2}A}{l}$
$=\frac{\mu }{{\mu }_0}={\mu }_r$
where, ${\mu }_r$ is the relative permeability of iron. Given,
$L=0.02mH,L^{\prime }=40mH$
$\because {\mu }_r=\frac{40}{0.02}=2000$